Band pass (2000 rad/s, 400 rad/s) from 3rd order

Frequency transformation

Frequency transformation is a change of the complex frequency variable that converts a normalized low-pass prototype filter (cutoff 1 rad/s) into a high-pass, band-pass or band-stop filter (or a low-pass with a different cutoff). The magnitude response of the prototype is mapped onto the new frequency axis, so the same attenuation values appear at the corresponding new frequencies.

Importance in filter design

• Only one set of approximation tables (low-pass poles and ladder element values) is needed; HP, BP and BS filters are derived from it.
• The difficult approximation step (finding $T(s)$ for given $\alpha_{max}$, $\alpha_{min}$) is done once, for the simpler low-pass case; specifications of the other types are converted to an equivalent LP specification.
• It can be applied either to the transfer function (substitute for $s$) or directly to each element of an LC ladder, so the circuit is obtained without re-deriving it.
• It reduces design time and errors, and keeps the properties of the prototype (equiripple, maximally flat, etc.).

Design of the band-pass filter

LP → BP: $s \to \dfrac{s^2+\omega_0^2}{Bs}$, with $\omega_0 = 2000$ rad/s and $B = 400$ rad/s.

• series $L$ → series $L' = L/B$, $C' = B/(L\omega_0^2)$;
• shunt $C$ → parallel $C' = C/B$, $L' = B/(C\omega_0^2)$.

Prototype from Table 1 ($n = 3$, $R_s = R_L = 1\ \Omega$): series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H.

$$\begin{aligned} L_1 = 1:&\quad L_1' = \frac{1}{400} = 2.5\ \text{mH},\quad C_1' = \frac{400}{1\times4\times10^6} = 100\ \mu\text{F} \\ C_2 = 2:&\quad C_2' = \frac{2}{400} = 5\ \text{mF},\quad L_2' = \frac{400}{2\times4\times10^6} = 50\ \mu\text{H} \\ L_3 = 1:&\quad L_3' = 2.5\ \text{mH},\quad C_3' = 100\ \mu\text{F} \end{aligned}$$

  Rs=1   [ Z1 ]          [ Z3 ]
 o-/\/\--[    ]---+------[    ]---+-----o
 |                |               |     +
 Vs            [ Y2 ]           RL=1    Vo
 |                |               |     -
 o----------------+---------------+-----o

Check: $1/\sqrt{2.5\text{ mH}\times100\ \mu\text{F}} = 2000$ rad/s. The 3 dB band edges are $\omega_{1,2} = \sqrt{\omega_0^2+(B/2)^2} \mp B/2 = 1810.0$ and $2210.0$ rad/s, so $\omega_2 - \omega_1 = 400$ rad/s. The result is a 6th-order band-pass filter.

These values are for $1\ \Omega$ terminations. For practical values, magnitude-scale by $k_m$ (e.g. $k_m = 1000$ for 1 kΩ terminations): multiply every L by $k_m$ and divide every C by $k_m$.